A disc of radius R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the centre of mass of the residual disc.
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Let ‘0’ be the origin of the system. R = radius of the smaller disc. 2R = radius of the bigger disc. The smaller disc is cut out from the bigger disc. As from the figure: 
$\text{m}_1=\pi\text{R}^2\text{T}\rho,\ \text{x}_1=\text{R},\ \text{y}_1=0$ $\text{m}_2=\pi(\text{2R})^2\text{T}\rho,\ \text{x}_2=0,\ \text{y}_2=0$ $\Big(\frac{-\pi\text{R}^2\text{T}\rho\text{R}+0}{\pi\text{R}^2\text{T}\rho+\pi(2\text{R})^2\text{T}\rho\text{R}},\frac{0+0}{\text{m}_1+\text{m}_2}\Big)$ $=\Big(\frac{-\pi\text{R}^2\text{T}\rho\text{R}}{3\pi\text{R}^2\text{T}\rho},0\Big)=\Big( -\frac{\text{R}}{3},0\Big)$ C.M. is at $\frac{\text{R}}3{}$ from the centre of bigger disc away from the centre of the hole.
$\text{m}_1=\pi\text{R}^2\text{T}\rho,\ \text{x}_1=\text{R},\ \text{y}_1=0$ $\text{m}_2=\pi(\text{2R})^2\text{T}\rho,\ \text{x}_2=0,\ \text{y}_2=0$ $\Big(\frac{-\pi\text{R}^2\text{T}\rho\text{R}+0}{\pi\text{R}^2\text{T}\rho+\pi(2\text{R})^2\text{T}\rho\text{R}},\frac{0+0}{\text{m}_1+\text{m}_2}\Big)$ $=\Big(\frac{-\pi\text{R}^2\text{T}\rho\text{R}}{3\pi\text{R}^2\text{T}\rho},0\Big)=\Big( -\frac{\text{R}}{3},0\Big)$ C.M. is at $\frac{\text{R}}3{}$ from the centre of bigger disc away from the centre of the hole.
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